Question

Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of OH− ions in the solution after the reaction has gone to completion. Assume that there is no volume change when adding the grams of NaOH

Answer 1

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Step-by-step explanation:

Moles of NaOH =
`(15.0 g)/(40 g/mol)=0.375 mol`

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

`Molarity=(Moles)/(Volume(L))`

`n=0.250 M* 0.150 L=0.0375 mol`

`NaOH+HNO_3\rightarrow NaNO_3+H_2O`

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

`(1)/(1)* 0.0375 mol` of NaOH

Moles of NaOH left unreacted in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

`NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)`

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

`1* 0.3375 moles=0.3375 mol` of hydroxide ion

The molarity of hydroxide ion in solution ;

`=(0.3375 mol)/(0.150 L)=2.25 M`

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.